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Sveta_85 [38]
2 years ago
7

If an air parcel is given a small push upward and it falls back to its original position, the atmosphere is said to be__________

.
Physics
1 answer:
nydimaria [60]2 years ago
6 0

If an air parcel is given a small push upward and it falls back to its original position, the atmosphere is said to be Stable.

A Stable atmospheric condition is a state where a body thrown like an parcel tends to return to its original position where it was present earlier even after being disturbed externally.

Stable conditions tends to oppose the forces that lead to the displacement of the object from original position.

When a parcel is given a small push upward and it falls back to its original position in air. This shows Stability in the atmosphere which bring back the sir parcel where it was present initially.

A body under the action of gravity in case of stable equilibrium conditions shows returning back of the body after sometime on reaching a certain height.

Learn more about Gravity here, brainly.com/question/4014727

#SPJ4

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Answer:

Q1)  Time taking by particle to travel the 40 km wrt. earth = 1.34\times10^{-6} sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = 1.285\times10^{-5} sec.

Explanation:

Given :

Speed of particle wrt. earth v=0.99537c

Distance between where particle is created and earth surface = 40 km

we know that,

⇒       v = \frac{x}{t}

Where x = 40\times10^{3} m, v = 0.99537c, we know speed of light c = 3 \times10^{8}

∴      t = \frac{x}{v}

         = \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }

      t = 1.34\times10^{-6} sec

∴ Thus, time taking by particle to travel the 40 km wrt. earth t = 1.34\times10^{-6} sec

According to the lorentz transformation,

⇒    l = l_{o} \sqrt{1-\frac{v^2}{c^2} }

Where l = improper length, l_{o} =proper length (distance measured wrt. rest frame) = 40 km

     l = 40 \sqrt{1-\frac{v^2}{c^2 }

     l = 40 \times 0.096

     l = 3.84 km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

   \Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }

Where \Delta t = improper time (wrt. earth frame time) =1.34\times10^{-6} sec ,  \Delta t _{o} = proper time (wrt. particle frame).

 1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}

 \Delta t_{o} = 1.285 \times10^{-5} sec

Thus, the time taking by particle to travel from where it is created to the surface of the earth = 1.285 \times10^{-5} sec.

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Answer: 10 s, 30 m/s , 150 m

Explanation:

Given

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The initial speed of a police motorcycle is u_2=0\ m/s

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